A vertical pole fixed to the horizontal ground is divided in the ratio 3 : 7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is:

Let us denote the foot of the pole on the ground by the point $$O$$. The pole itself is vertical, so its topmost point is directly above $$O$$.

The mark that divides the pole is such that the lower part and the upper part are in the ratio $$3:7$$, with the lower part shorter. If we let the common proportionality constant be $$k$$ metres, then

$$\text{length of lower part } = 3k,$$ $$\text{length of upper part } = 7k,$$ $$\text{total height of pole } = 3k + 7k = 10k.$$

Now choose a point $$P$$ on the horizontal ground such that the horizontal distance between $$P$$ and the foot $$O$$ of the pole is $$18\text{ m}$$. According to the question, the two parts of the pole individually subtend equal visual angles at the point $$P$$.

It is convenient to introduce the heights of the relevant points above the ground:

$$A \; (=O) : \text{height } 0,$$ $$B : \text{height } 3k,$$ $$C : \text{height } 10k.$$

The two visual angles at $$P$$ are:

$$\theta_1 = \angle APB = \arctan\!\left(\frac{3k}{18}\right) - \arctan\!\left(\frac{0}{18}\right)$$ $$\phantom{\theta_1}= \arctan\!\left(\frac{3k}{18}\right),$$

$$\theta_2 = \angle BPC = \arctan\!\left(\frac{10k}{18}\right) - \arctan\!\left(\frac{3k}{18}\right).$$

The condition given is $$\theta_1 = \theta_2$$, so

$$\arctan\!\left(\frac{3k}{18}\right) \;=\; \arctan\!\left(\frac{10k}{18}\right) - \arctan\!\left(\frac{3k}{18}\right).$$

Set $$\alpha = \arctan\!\left(\frac{3k}{18}\right).$$ Then the equality becomes

$$\alpha = \arctan\!\left(\frac{10k}{18}\right) - \alpha,$$

so that

$$2\alpha = \arctan\!\left(\frac{10k}{18}\right).$$

Next we use the double-angle formula for tangent, namely $$\tan(2\alpha) = \frac{2\tan\alpha}{1 - \tan^{2}\alpha}.$$

We have $$\tan\alpha = \frac{3k}{18} = \frac{k}{6},$$ hence

$$\tan(2\alpha) = \frac{2\left(\tfrac{k}{6}\right)}{1 - \left(\tfrac{k}{6}\right)^{2}} = \frac{\tfrac{2k}{6}}{1 - \tfrac{k^{2}}{36}} = \frac{\tfrac{k}{3}}{1 - \tfrac{k^{2}}{36}}.$$

But $$\tan(2\alpha)$$ is also, from above, $$\tan\!\Bigl(\arctan\!\bigl(\tfrac{10k}{18}\bigr)\Bigr) = \frac{10k}{18} = \frac{5k}{9}.$$

Equating the two expressions for $$\tan(2\alpha)$$ gives

$$\frac{\tfrac{k}{3}}{1 - \tfrac{k^{2}}{36}} = \frac{5k}{9}.$$

The factor $$k\neq 0$$ cancels out, leaving

$$\frac{1}{3\!\!}\;\Big/\!\!\left(1 - \frac{k^{2}}{36}\right) = \frac{5}{9}.$$

Cross-multiplying,

$$\frac{1}{3} = \frac{5}{9}\left(1 - \frac{k^{2}}{36}\right).$$

Multiply both sides by $$9$$:

$$3 = 5\left(1 - \frac{k^{2}}{36}\right).$$

Expand the right-hand side:

$$3 = 5 - \frac{5k^{2}}{36}.$$

Transpose $$5$$ to the left:

$$3 - 5 = -\frac{5k^{2}}{36},$$ $$-2 = -\frac{5k^{2}}{36}.$$

Multiply both sides by $$-1$$:

$$2 = \frac{5k^{2}}{36}.$$

Multiply by $$36$$:

$$72 = 5k^{2}.$$

Hence

$$k^{2} = \frac{72}{5}, \quad k = \sqrt{\frac{72}{5}} = \frac{6\sqrt{2}}{\sqrt{5}} = \frac{6\sqrt{10}}{5}.$$

The total height of the pole is

$$10k = 10 \times \frac{6\sqrt{10}}{5} = \frac{60\sqrt{10}}{5} = 12\sqrt{10}\text{ metres}.$$

Hence, the correct answer is Option C.